This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

Fig. 226. Resultant Forces Acting on Voussoirs of a Full-Centered Arch.

We shall assume that the arch carries a railroad track and a heavy class of traffic. The weight of roadbed and track may be computed as follows: The ties are to be 8 feet long; the weight of the roadbed and track (and also the live load) is assumed to be distributed over an area 8 feet wide.

Two rails at 100 pounds per yard will weigh, per square foot of surface..... | 8.4 | lbs. |

Oak ties, weighing 150 pounds per tie, will weigh, per square foot of surface...... | 9.4 | ,, |

Weight of ballast at 100 pounds per cubic foot, average depth 9 inches.... | 75.0 | ,, |

Total weight | 92.8 | lbs. |

This is the equivalent of 0.58 foot depth of stone, and we therefore add this uniform depth to the reduced load line for the earth.

A 50-ton freight-car, fully loaded, will weigh 134,000 pounds; with a length between bumpers of 37 feet, this will exert a pressure of about 450 pounds per square foot on a strip 8 feet wide. This is equivalent to 2.8 feet of masonry. We shall therefore consider this as a requirement for uniform loading over the whole arch.

It would be more precise to consider the actual wheel loads for the end trucks of two such cars which are immediately following each other; but since the effect of this would be even less than that of the calculation for a locomotive, which will be given later, and since the deep cushion of earth filling will largely obliterate the effect of concentrated loads, the method of considering the loading as uniformly distributed will be used. We therefore add the uniform ordinate equal to 2.8 feet over the whole arch. We shall call this the first condition of loading.

We shall assume for the concentrated loading, a consolidation locomotive with 40,000 pounds on each of the four driving axles, spaced 5 feet apart. This means a wheel base 15 feet long; and we shall assume that this extends over voussoirs 1 to 8 inclusive, while the loading of 450 pounds per square foot is on the other portion of the arch. A weight of 40,000 pounds on an axle, which is supposed to be distributed over an area 5 feet long and 8 feet wide, gives a pressure of 1,000 pounds per square foot, or it would add an ordinate of G.33 feet of stone; these ordinates are added above the load line representing the load of the roadbed and track. We shall call this the second condition of loading.

The load for each voussoir is determined by the method given in Article 417. The direction of the pressure on the voussoir is determined by drawing a line toward the extrados center from the intersection of the vertical through the trapezoid of loading with the extrados. The length of that vertical is laid off below that point of intersection; then a horizontal line drawn from the lower end of the vertical intersects the line of force at a point which measures (he amount of that pressure on the voussoir. The area of the voussoir is determined as described in Article 417; and the resultant of the loading and the weight of the voussoir is obtained. This is indicated as force No. 1 in Fig. 226. In this case, it includes the locomotive loading on the left-hand side of the arch. The forces acting on voussoirs Nos. 2, 3, 4, 5, 6, 7, and 8 are similarly determined. The forces on voussoirs Nos. 9 to 17 inclusive, on the basis of the uniformly distributed load equal to 450 pounds per square foot, are also similarly determined. The loads on voussoirs Nos. 10 to 17 inclusive will be considered to measure the loads on voussoirs Nos. 8 to 1 inclusive, for the first condition of loading. The loading with the locomotive over voussoirs Nos. 1 to 8, and cars over voussoirs Nos. 9 to 17, constitutes the second condition of loading.

As described above, the arrows representing the forces in Fig. 226 are drawn at a scale such that each 3/8 of an inch represents 2 cubic feet of masonry, or 320 pounds; therefore every inch will represent the quotient of 320 divided by 3/8, or 853 pounds per linear inch. The practical method of making a scale for this use is illustrated in the diagram in the upper right-hand corner of Fig. 226. We may draw a horizontal line as a scale line, and lay off on it, with a decimal scale, a length ca which represents, at some convenient scale, a length of 800. Drawing the line ab at any convenient angle, we lay off from the point c the length cb to represent 853 at the same scale as that used for ca. The line cd is then laid off to represent 7,000 units at the scale of 800 units per inch. By drawing a line from d parallel to ha, we have the distance ce, which represents 7,000 units at the scale of 853 units per inch. By trial, a pair of dividers may be so spaced that it steps off precisely seven equal parts for the distance ce; or the line ce may also be divided into equal parts by laying off on cd to the decimal scale, the seven equal parts of 1,000 each which are at the scale of 800 units per inch; and then lines may be drawn from these points parallel to ba and de. The last division may be similarly divided into 10 equal parts, which will represent 100 pounds each.

Using dividers, the resultant force on each voussoir from No. 1 to No. 17 may be scaled off as follows:

1 | 7,825 |

2 | 5,970 |

3 | 4,940 |

4 | 4,190 |

5 | 3,725 |

6 | 3,380 |

7 | 3,170 |

8 | 3,040 |

9 | 1,880 |

10 | 1,910 |

11 | 2,040 |

12 | 2,200 |

13 | 2,400 |

14 | 2,905 |

15 | 3,570 |

16 | 4,420 |

17 | 6,005 |

The student should note the three dotted curves in the lower part of the figure, which have been drawn through the extremities of the forces. The only object in drawing these three curves is merely to note the uniformity with which the ends of these arrows form a regular curve. If it was found that one of the forces did not pass through this curve, it would probably imply a blunder in the method of determining that particular force. Even if such curves are not actually drawn in, it is well to observe that the points do come on a regular curve, as this constitutes one of the checks on the graphical solution of problems.

Fig. 226 is merely the beginning of the problem of determining the stresses in the arch. In order to save the complication of the figure, the arch itself and the resultant forces (1 to 17) are repeated in Fig. 227, the direction, intensity, and point of application of these forces being copied from one figure to the other.

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